How do you differentiate $f(x) = sin^2 x + 1/2 cot x-tanx$ ?

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Answer 1 · 2018-06-01T17:56:25

$f'(x)=sin(2x)-1/(2sin^2(x))-1/cos^2(x)$

Writing
$f(x)=sin^2(x)+1/2*cos(x)/sin(x)-sin(x)/cos(x)$

$f'(x)=2sin(x)cos(x)+1/2*(-sin^2(x)-cos^2(x))/sin^2(x)-(sin^2(x)+cos^2(x))/cos^2(x)$

so

$f'(x)=sin(2x)-1/(2*sin^2(x))-1/cos^2(x)$

How do you differentiate f(x) = sin^2 x + 1/2 cot x-tanxf(x) = sin^2 x + 1/2 cot x-tanx?